Answer to Question #342780 in Statistics and Probability for elay

Question #342780

An auto battery company claims that their batteries’ mean

life is 50 months. In order to check this claim, a DTI

researcher took a random sample of 18 of these batteries

and found that the mean life is 48.8 months with a

standard deviation of 7 months. Assume that battery life

follows a normal distribution, test with 90% confidence

whether the companies’ claim is different from the true


1
Expert's answer
2022-05-20T06:45:36-0400

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=50"

"H_a:\\mu\\not=50"

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is "\\alpha = 0.10," "df=n-1=17" degrees of freedom, and the critical value for a two-tailed test is "t_c =1.739607."The rejection region for this two-tailed test is "R = \\{t:|t|>1.739607\\}."

The t-statistic is computed as follows:



"t=\\dfrac{48.8-50}{7\/\\sqrt{18}}=-0.7273"

Since it is observed that "|t| = 0.7273<1.739607=t_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed "df=17" degrees of freedom, "t=-0.7273" is "p= 0.47694," and since "p= 0.47694>0.10=\\alpha," it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean "\\mu" is different than 50, at the "\\alpha = 0.10" significance level.


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